Question

Prove $1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$

Answer 1

$1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$
For n = 1
$P\left(1\right)=(2\times 1-1{)}^2=\frac{1(2\times 1-1)(2\times 1+1)}{3}\Rightarrow 1=\frac{1\times 1\times 3}{3}$
$\therefore$ P (1) is true
Let P (n) be true for n = k
$\therefore P\left(k\right)=1^2+3^2+5^2+\cdots +(2k-1{)}^2=\frac{k(2k-1)(2k+1)}{3}\text{For n = k + 1}\text{R.H.S.}=\frac{(k+1)(2k+1)(2k+3)}{3}L.H.S.=\frac{k(2k-1)(2k+1)}{3}+(2k+1{)}^2=(2k+1)[\frac{k(2k-1)}{3}+(2k+1\left)\right]=(2k+1)\left[\frac{2k^2-k+6k+3}{3}\right]=\frac{(2k+1)(2k^2+5k+3)}{3}=\frac{(2k+1)(k+1)(2k+3)}{3}=\frac{(k+1)(2k+1)(2k+3)}{3}\therefore P(k+1)\text{is true}$
Thus P(k) is true $\Rightarrow$ P(k + 1) is true
Hence by principle fo mathematical induction,